burnside's problem, spanning trees and tilings

mspGeometry & Topology 18 (2014) 179–210

Burnside’s Problem, spanning trees and tilings

BRANDON SEWARD

In this paper we study geometric versions of Burnside’s Problem and the von Neu-mann Conjecture. This is done by considering the notion of a translation-like action.Translation-like actions were introduced by Kevin Whyte as a geometric analogueof subgroup containment. Whyte proved a geometric version of the von NeumannConjecture by showing that a finitely generated group is nonamenable if and only if itadmits a translation-like action by any (equivalently every) nonabelian free group. Westrengthen Whyte’s result by proving that this translation-like action can be chosento be transitive when the acting free group is finitely generated. We furthermoreprove that the geometric version of Burnside’s Problem holds true. That is, everyfinitely generated infinite group admits a translation-like action by Z . This answersa question posed by Whyte. In pursuit of these results we discover an interestingproperty of Cayley graphs: every finitely generated infinite group G has some locallyfinite Cayley graph having a regular spanning tree. This regular spanning tree can bechosen to have degree 2 (and hence be a bi-infinite Hamiltonian path) if and only ifG has finitely many ends, and it can be chosen to have any degree greater than 2 ifand only if G is nonamenable. We use this last result to then study tilings of groups.We define a general notion of polytilings and extend the notion of MT groups and cccgroups to the setting of polytilings. We prove that every countable group is poly-MTand every finitely generated group is poly-ccc.

20F65; 05C25, 05C63

1 Introduction

This paper focuses on the notion of a translation-like action, introduced by Kevin Whytein [14], and its relevance to geometric versions of Burnside’s Problem and the vonNeumann Conjecture. The classical Burnside’s Problem asks if every finitely generatedinfinite group contains Z as a subgroup, and the classical von Neumann Conjecturestates that a group is nonamenable if and only if it contains a nonabelian free group asa subgroup. Although both of these problems are known to have negative answers (bywork of Golod and Shafarevich [7] and Olshanskii [9], respectively), translation-likeactions provide us with a new geometric perspective of these classical problems. Studyof this notion in turn leads us to findings about Hamiltonian paths, regular spanning

Published: 23 January 2014 DOI: 10.2140/gt.2014.18.179

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180 Brandon Seward

trees of Cayley graphs and tilings of groups. While we are primarily interested intranslation-like actions on groups, we are naturally led to consider translation-likeactions on graphs as well. We therefore present a general definition of translation-likeactions.

Definition 1.1 (Whyte [14]) Let H be a group and let .X; d/ be a metric space. Aright action � of H on X is translation-like if it satisfies the following two conditions:

(i) The action is free (ie x � hD x implies hD 1H ).

(ii) For every h 2H the set fd.x;x � h/ j x 2X g is bounded.

We view every finitely generated group as a metric space by using a left-invariant wordlength metric associated to some finite generating set, and we view every connectedgraph as a metric space by using the path length metric (see the next section for details).

We point out that if H acts translation-like on a finitely generated group or a graph X,then for all h 2H the map x 2X 7! x�h 2X is bilipschitz [14]. Thus, in the contextof finitely generated groups and graphs one could equivalently define translation-likeactions to be free actions by bilipschitz maps at bounded distance from the identitymap.

Whyte’s motivation in defining translation-like actions is that it serves as a geometricanalogue of subgroup containment. Specifically, we have the following propositionwhose proof is trivial.

Proposition 1.2 (Whyte [14]) Let G be a finitely generated group and let H � G.Then the natural right action of H on G is a translation-like action.

In [14], Whyte suggested that conjectures relating geometric properties to subgroupcontainment may hold true if one requires a translation-like action in place of a subgroup.Specifically, Whyte mentioned the following three well known conjectures/problems.

(1) Burnside’s Problem: Does every finitely generated infinite group contain Z as asubgroup?

(2) The von Neumann Conjecture: A group is nonamenable if and only if it containsa nonabelian free subgroup.

(3) Gersten Conjecture: A finitely generated group is not word hyperbolic if andonly if it contains some Baumslag–Solitar group.

All three of the above conjectures/problems were answered negatively by Golod andShafarevich [7], Olshanskii [9] and Brady [3], respectively. Whyte suggested thefollowing “geometric reformulations” of these problems.

Geometry & Topology, Volume 18 (2014)

Burnside’s Problem, spanning trees and tilings 181

(1 0 ) Geometric Burnside’s Problem: Does every finitely generated infinite groupadmit a translation-like action by Z?

(2 0 ) Geometric von Neumann Conjecture: A finitely generated group is nonamenableif and only if it admits a translation-like action by a nonabelian free group.

(3 0 ) Geometric Gersten Conjecture: A finitely generated group is not word hyperbolicif and only if it admits a translation-like action by some Baumslag–Solitar group.

Whyte proved (2 0 ).

Theorem 1.3 (Geometric von Neumann Conjecture; Whyte [14]) A finitely gener-ated group is nonamenable if and only if it admits a translation-like action by some(equivalently every) nonabelian free group.

In this paper we answer (1 0 ) in the affirmative.

Theorem 1.4 (Geometric Burnside’s Problem) Every finitely generated infinite groupadmits a translation-like action by Z.

We actually prove something stronger than both (1 0 ) and (2 0 ). The main distinction ofthe following theorem, as opposed to Theorems 1.3 and 1.4, is the transitivity of theaction.

Theorem 1.5 Let G be a finitely generated infinite group. Then:

(i) G has finitely many ends if and only if it admits a transitive translation-likeaction by Z.

(ii) G is nonamenable if and only if it admits a transitive translation-like action byevery finitely generated nonabelian free group.

In particular, every finitely generated infinite group admits a transitive translation-likeaction by some (possibly cyclic) free group.

In pursuit of the above theorem we prove two graph-theoretic results.

Theorem 1.6 Let � be a connected graph whose vertices have uniformly boundeddegree. Then � is bilipschitz equivalent to a graph admitting a Hamiltonian path if andonly if � has at most two ends.

Theorem 1.7 Let ƒ1 and ƒ2 be two trees. If every vertex of ƒ1 and ƒ2 has degreeat least three and if the vertices of ƒ1 and ƒ2 have uniformly bounded degree, then ƒ1

and ƒ2 are bilipschitz equivalent.


182 Brandon Seward

In [11], Papasoglu proved Theorem 1.7 under the additional assumption that ƒ1 and ƒ2

are regular trees of degree at least 4.

Theorem 1.5 above may be stated in an equivalent form without mention of translation-like actions. This alternative form, below, illustrates a peculiar property of the set ofCayley graphs associated to a group. The Cayley graphs mentioned in the followingtheorem are understood to be Cayley graphs coming from finite generating sets.

Theorem 1.8 If G is a finitely generated infinite group then G has a Cayley graphadmitting a regular spanning tree. In fact, for every integer k > 2 the following hold:

(i) G has finitely many ends if and only if G has a Cayley graph admitting aHamiltonian path (ie a regular spanning tree of degree 2).

(ii) G is nonamenable if and only if G has a Cayley graph admitting a regularspanning tree of degree k .

We give an example to show that “a Cayley graph” cannot be replaced by “every Cayleygraph.” Thus only certain Cayley graphs have regular spanning trees, but yet everyfinitely generated infinite group has at least one Cayley graph with this property.

We give a nontrivial application of Theorem 1.8 to tilings of groups, the final topicof this paper. Tilings of groups have been studied by Chou [5], Gao, Jackson andSeward [6] and Weiss [13] due to their usefulness in studying group actions, dynamics,equivalence relations and general marker structures. From the algebraic and geometricviewpoints, tilings of groups are still rather mysterious as they have received littleinvestigation and much is still unknown. In this paper we study polytiles and polytilings.These are more general notions than studied in [5; 6; 13].

Definition 1.9 For a countable group G, a tuple .T1;T2; : : : ;Tk/ of finite subsetsof G, all containing the identity element, is a polytile if there are nonempty sets�1; �2; : : : ; �k �G so that G is the disjoint union

G Da

1�i�k; ı2�i

ıTi :

We call the tuple P D .�1; : : : ; �k IT1; : : : ;Tk/ a polytiling of G. The partition of G

expressed above is called the partition of G induced by P. In the case k D 1, polytilesare referred to as monotiles and polytilings are referred to as monotilings.

Chou [5] and Weiss [13] studied which groups G have the property that for every finiteF �G there is a monotile T with F � T . Weiss called groups with this property MTgroups (an acronym for “monotileable”). Chou [5] proved that residually finite groups,



elementary amenable groups (in particular solvable groups) and groups which are freeproducts of nontrivial groups are MT. Beyond these results, no other groups are knownto be MT. Interestingly, there are also no known examples of groups which are notMT. In this paper we define and consider a related weaker question: Which groups arepoly-MT? Of course, care needs to be taken in defining a notion of poly-MT as anytuple .T1;T2; : : : ;Tk/ containing a singleton is a polytile. To rule out such trivialities,one wants to consider tuples .T1;T2; : : : ;Tk/ where the cardinalities of the Ti donot vary too much. The following definition seems to give the strongest restriction inthis sense.

Definition 1.10 A polytile .T1;T2; : : : ;Tk/ is fair if all of the Ti have the samenumber of elements. Similarly, a polytiling .�1; : : : ; �k IT1; : : : ;Tk/ is fair if thepolytile .T1; : : : ;Tk/ is fair.

Definition 1.11 A countable group G is poly-MT if for every finite F �G there is afair polytile .T1; : : : ;Tk/ with F � T1 .

We prove the following. Recall that a group is locally finite if every finite subsetgenerates a finite subgroup.

Theorem 1.12 Every countable group is poly-MT. Furthermore, if G is a countablyinfinite nonlocally finite group then for every finite F � G and all sufficiently largen 2N there is a fair polytile .T1; : : : ;Tk/ with F � T1 and jT1j D n.

In addition to studying individual tilings, as Chou and Weiss did, we also studysequences of tilings as was done by Gao, Jackson and Seward in [6]. The followingdefinition generalizes definitions appearing in [6] to the context of polytilings.

Definition 1.13 Let G be a countable group, and let

.Pn/n2N D .�n1; : : : ; �

nk.n/IT

n1 ; : : : ;T

nk.n//n2N

be a sequence of polytilings. This sequence is fair if each of the Pn is fair. Thesequence is said to be coherent if for each n 2N the partition of G induced by Pn isfiner than the partition of G induced by PnC1 . The sequence is said to be centered if1G 2�

n1

for all n 2N . Finally, the sequence is cofinal if T n1� T nC1

1for all n and

G DS

n2NT n1

. The three adjectives “centered, cofinal and coherent” are abbreviatedto ccc.


184 Brandon Seward

We point out that if a sequence of polytilings .Pn/n2N is ccc then every two elementsg1;g2 2 G lie in the same class of the partition induced by Pn for all sufficientlylarge n. Also notice that a group is MT (poly-MT) if and only if it admits a cofinalsequence of monotilings (fair polytilings). Thus having a ccc sequence of monotilings(fair polytilings) is stronger than being MT (poly-MT).

In [6], Gao, Jackson and Seward studied which groups admit a ccc sequence ofmonotilings. They called such groups ccc groups. They proved that the followinggroups are ccc groups: locally finite groups; residually finite groups; nilpotent groups;solvable groups G in which ŒG;G� is polycyclic; and groups which are free products ofnontrivial groups. These are currently the only groups known to be ccc, and furthermorethere are no known examples of groups which are not ccc. We consider the weakercondition of which groups are poly-ccc.

Definition 1.14 A countable group G is poly-ccc if it admits a ccc sequence of fairpolytilings.

Notice that just as ccc is a stronger condition than MT, poly-ccc is a stronger conditionthan poly-MT. We prove the following.

Theorem 1.15 Every finitely generated group is poly-ccc. That is, every finitelygenerated group admits a ccc sequence of fair polytilings.

Interestingly, we do not know if countable nonfinitely generated groups are poly-ccc.

We point out that the proofs of these results on tilings rely heavily on the fact that everyfinitely generated infinite group has some Cayley graph having a regular spanning tree(Theorem 1.8). These theorems on tilings therefore demonstrate an application of thisresult.

The organization of the remainder of the paper is as follows. In Section 2, we presentformal definitions, notation and some simple observations. In Section 3, we studytranslation-like actions of Z which are transitive and we also study bi-infinite Hamil-tonian paths on graphs. In this section we prove Theorem 1.6. We then use thesefindings in Section 4 to prove Theorem 1.4 (the Geometric Burnside’s Problem) as wellas the first clause of both Theorems 1.5 and 1.8. Section 5 is devoted to strengtheningthe Geometric von Neumann Conjecture as well as constructing Cayley graphs havingregular spanning trees. In this section we prove Theorem 1.7 and prove the remainingstatements of Theorems 1.5 and 1.8. Finally, in Section 6 we study tilings of groupsand prove Theorems 1.12 and 1.15.



Acknowledgements This research was supported by a National Science FoundationGraduate Research Fellowship. The author would like to thank Khalid Bou-Rabee forhelpful comments regarding an earlier version of this paper.

2 Preliminaries

In this section we go over some basic definitions, observations and notation. A fun-damental notion which will be used frequently is that of a bilipschitz map. If .X; d/and .Y; �/ are two metric spaces and f W X ! Y is a function, then f is Lipschitz ifthere is a constant c > 0 such that for all x1;x2 2X,

�.f .x1/; f .x2//� c � d.x1;x2/:

If we additionally have that

1c� d.x1;x2/� �.f .x1/; f .x2//

for all x1;x2 2X, then f is called bilipschitz. The metric spaces .X; d/ and .Y; �/are bilipschitz equivalent if there is a bijective bilipschitz function f W X ! Y . Finally,if d and � are two metrics on X, then they are called bilipschitz equivalent if theidentity map .X; d/! .X; �/ is bilipschitz.

Notice that if d and � are bilipschitz equivalent metrics on X then an action of H

on X is translation-like with respect to the metric d if and only if it is translation-likewith respect to the metric � . Thus to discuss translation-like actions on X, it sufficesto specify a family of bilipschitz equivalent metrics on X. This is particularly usefulwhen X DG is a finitely generated group. If G is a finitely generated group, then wecan specify a natural family of bilipschitz equivalent metrics on G. Let S be any finitegenerating set for G, and let dS be the metric on G given by

dS .g; h/Dminfn 2N j g�1h 2 .S [S�1/ng

(where A0 D f1Gg and An D fa1 � a2 � � � an j ai 2 Ag for A � G ). Notice dS

is left-invariant, which means that dS .kg; kh/ D dS .g; h/ for all k;g; h 2 G. Wecall dS the left-invariant word length metric corresponding to S . All of the metricsfdS j S �G is a finite generating setg are bilipschitz equivalent. To see this, one onlyneeds to consider expressing the elements of one generating set as products of elementsof the other generating set. Whenever discussing translation-like actions on a finitelygenerated group G, we will always use the family of bilipschitz equivalent metrics justspecified.

Most of our arguments in this paper rely on studying the structure of various graphs.We therefore review some notation and terminology related to graphs. Let � be a


186 Brandon Seward

graph. We denote the vertex set of � by V.�/ and the edge set of � by E.�/. Thegraph � is regular if every vertex has the same degree, and it is even if the degree ofevery vertex is finite and even. A subgraph of � is a graph ˆ where V.ˆ/ � V.�/and E.ˆ/� E.�/. We will write ˆ� � to denote that ˆ is a subgraph of �. We saya subgraph ˆ of � is spanning if V.ˆ/ D V.�/. A path P in � is a sequence ofvertices of � such that vertices which are consecutive in the sequence are joined by anedge in �. Formally we represent P as a function from a (finite or infinite) subintervalof Z into V.�/. If there is a smallest a 2 dom.P / then we call P .a/ the initial vertexor starting vertex of P. If there is a largest b 2 dom.P / then we call P .b/ the finalvertex or ending vertex of P. If v 2 V.�/, then we say that P traverses v or visits vif there is i 2 dom.P / with P .i/D v . If .v0; v1/ 2 E.�/, we say that P traverses theedge .v0; v1/ if there is i 2Z with i; iC12 dom.P / and fP .i/;P .iC1/gD fv0; v1g.A Hamiltonian path is a path which visits every vertex precisely one time. These pathswill be particularly important to us. A similar notion is that of an Eulerian path. AnEulerian path is a path which traverses each edge precisely one time. We realize thevertex set of every connected graph � as a metric space by using the path length metric.Specifically, the distance between u; v 2 V.�/ is the infimum of the lengths of pathsjoining u and v . If no metric is specified on a graph, then it is understood that we areusing the path length metric. We say that two graphs are bilipschitz equivalent if theirvertex sets equipped with their respective path length metrics are bilipschitz equivalent.Finally, if A� V.�/ and k � 1, we write N�

k.A/ to denote the union of A with the

set of all vertices v 2 V.�/ which are within distance k of some vertex of A.

In this paper we will discuss translation-like actions not only on groups but also onconnected graphs as well (using the path length metric described in the previousparagraph). Many times these graphs will arise in the form of Cayley graphs. If G is afinitely generated group and S is a finite generating set for G, then the right Cayleygraph of G with respect to S , denoted Cay.GIS/, is the graph with vertex set G

and edge relation f.g;gs/ j g 2 G; s 2 S [ S�1g. There is a symmetric notion ofa left Cayley graph where the edge set is f.g; sg/ j g 2 G; s 2 S [ S�1g. The leftand right Cayley graphs are graph isomorphic via the map g 7! g�1. In this paperwe will always use the term Cayley graph to mean a right Cayley graph Cay.GIS/where S is a finite generating set for G (we never discuss or consider Cayley graphscorresponding to infinite generating sets). If G is a finitely generated group and S �G

is a finite generating set, then there are two corresponding metrics on G : dS (describedearlier in this section) and the path length metric coming from Cay.GIS/. It is easyto see that these two metrics are identical. Thus we can freely and without concernswitch between discussing translation-like actions on G and translation-like actions ona Cayley graph of G.



Finally, we review the notion of the number of ends of a graph and of a finitelygenerated group. For a connected graph �, the number of ends of � is defined to bethe supremum of the number of infinite connected components of � �A as A rangesover all finite subsets of E.�/. If G is a finitely generated group, then it is known thatall Cayley graphs of G have the same number of ends (see for example Bridson andHaefliger [4, Section I.8, Proposition 8.29]). Thus the number of ends of G is definedto be the number of ends of any Cayley graph of G. Finite groups have 0 ends, andfinitely generated infinite groups have either 1, 2 or infinitely many ends [4, Section I.8,Theorem 8.32].

3 Hamiltonian paths

In this section we focus on actions of Z on graphs which are both translation-like andtransitive. The existence of such actions has a nice graph theoretic characterization, asthe following lemma shows.

Lemma 3.1 A graph � admits a transitive translation-like action by Z if and onlyif � is bilipschitz equivalent to a graph admitting a bi-infinite Hamiltonian path.

Proof First suppose there is a translation-like action � of Z on � which is transitive.Let d be the path length metric on � and let n� 1 be such that d.v; v �1/� n for allv 2V.�/ (notice that v �1¤ v , but rather v �0D v ). Let � 0 be the graph with vertexset V.�/ and let there be an edge between v0 and v1 if and only if there is a path in �of length at most n joining v0 to v1 . Let d 0 be the path length metric on � 0 . Then forv0; v1 2 V.�/,

1nd.v0; v1/� d 0.v0; v1/� d.v0; v1/:

Thus the identity map idW V.�/!V.� 0/ is a bilipschitz bijection between � and � 0 . Ifwe fix v 2V.� 0/ and define P W Z!V.� 0/ by P .k/D v�k , then P is a Hamiltonianpath on � 0 (since the action of Z is free and transitive).

Now suppose that there is a graph � 0 , a Hamiltonian path P W Z!� 0 , and a bilipschitzbijection �W V.�/! V.� 0/. First define an action of Z on � 0 by setting

v � nD P .nCP�1.v//

for n 2 Z and v 2 V.� 0/. This action is free and transitive since P is a Hamiltonianpath. Also, it is clear that v � n is at most distance n from v . Therefore this is atransitive translation-like action of Z on � 0 . Now we define a transitive translation-likeaction of Z on � as follows. For n 2 Z and v 2 V.�/ set

v � nD ��1.�.v/� n/:


188 Brandon Seward

Since � is bijective, this action is free and transitive. Also, since � is bilipschitz and�.v/� n is distance at most n from �.v/, it follows that the distance between v � n

and v is bounded independently of v 2 V.�/. Thus this is a transitive translation-likeaction.

In the case of Cayley graphs the previous lemma takes the following more appealingform.

Corollary 3.2 A finitely generated group G admits a transitive translation-like actionby Z if and only if G has a Cayley graph admitting a bi-infinite Hamiltonian path.

Proof If G admits a transitive translation-like action by Z, then let SDS�1 be a finitegenerating set for G and let Cay.GIS/ be the corresponding Cayley graph. Follow theproof of Lemma 3.1 to get a graph � 0 having a Hamiltonian path. Now notice that � 0 isgraph isomorphic to a Cayley graph Cay.GIT /, where T DS[S2[� � �[Sn for somen 2 N . Conversely, if Cay.GIT / is a Cayley graph of G which has a Hamiltonianpath, then Cay.GIT / is bilipschitz equivalent to itself and so by Lemma 3.1 admits atransitive translation-like action by Z. This clearly is also a transitive translation-likeaction of Z on G.

We now give a general sufficient condition for a graph to be bilipschitz equivalent to agraph admitting a bi-infinite Hamiltonian path.

Theorem 3.3 Let � be a connected infinite graph whose vertices have uniformlybounded degree. Then � is bilipschitz equivalent to a graph admitting a bi-infiniteHamiltonian path if and only if � has at most two ends.

Proof First suppose that � is bilipschitz equivalent to � 0 and that � 0 has a bi-infiniteHamiltonian path. Then � 0 contains the canonical Cayley graph of Z as a spanningsubgraph. Any spanning subgraph must have at least as many ends as the originalgraph. So this implies that � 0 has at most two ends. Since the number of ends of agraph is preserved by bilipschitz equivalence [4, Section I.8, Proposition 8.29], � hasat most two ends.

Now suppose that � has at most two ends. Notice that � cannot have 0 ends since �is infinite, connected and locally finite. So � has either 1 end or 2 ends. We willfirst construct a path on � which will be very similar to a bi-infinite Eulerian path.The path we construct will visit every vertex at least once and traverse every edgeat most twice. In the construction we will use a theorem of Erdos, Grünwald andWeiszfeld [2, Section I.3, Theorem 14] which says that if ƒ is a countably infiniteconnected multigraph then ƒ admits a bi-infinite Eulerian path if and only if thefollowing conditions are satisfied:



(i) If the degree of a vertex is finite then it is even.

(ii) ƒ has at most 2 ends.

(iii) If ˆ is a finite, even subgraph of ƒ then ƒ�E.ˆ/ has only one infinite connectedcomponent.

Recall that a multigraph is a graph where loops are allowed and multiple edges betweenvertices are allowed.

First suppose that � has 1 end. Let � 0 be the multigraph obtained from � by doublingeach of the edges. In other words, V.� 0/D V.�/ and for each u; v 2 V.� 0/ � 0 hastwice as many edges joining u and v as � does. Then � 0 is an infinite connectedone-ended multigraph and every vertex of � 0 has even degree. Thus there is an Eulerianpath P W Z! V.� 0/D V.�/. Clearly P is a path on � which traverses every edgetwice.

Now suppose that � has two ends. We claim that there is a finite set E � E.�/ suchthat the graph � �E has precisely two connected components with both componentsinfinite. Since � has two ends, there is a finite set F � E.�/ such that � �F hastwo infinite connected components. We wish to shrink F in order to connect the finitecomponents to the infinite components, without connecting the two infinite componentsto one another. We do this in two steps. First, let C be one of the infinite connectedcomponents of ��F, and let F 0 be the set of those edges in F which have an endpointin C. Now � �F 0 still has two infinite connected components, one of which is C.The advantage we now have is that every edge in F 0 has an endpoint in C. We wantto connect the finite connected components of � �F 0 to C without connecting C tothe other infinite connected component of � �F 0. So for the second step let E be theset of those edges in F 0 which have both endpoints contained in infinite connectedcomponents of � �F 0. Then � �E has no finite connected components but it has twoinfinite connected components. This proves the claim as E has the desired property.

Let E � E.�/ be as in the previous paragraph. Let C1 and C2 be the two connectedcomponents of � �E . So C1 and C2 are infinite. Since � is connected, there isp 2 C1 which is adjacent to C2 in �. Let ƒ1 and ƒ2 be the induced subgraphson C1 and C2 [ fpg, respectively. Specifically, V.ƒ1/ D C1 , V.ƒ2/ D C2 [ fpg

and u; v 2 V.ƒi/ are joined by an edge if and only if they are joined by an edgein �. For i D 1; 2, let Wi W N ! V.ƒi/ be a path that begins at p and does notself-intersect (ie Wi is a path beginning at p and going off to infinity in ƒi ). Suchpaths exist since ƒ1 and ƒ2 are infinite, connected and locally finite. For i D 1; 2,let ƒ0i be the multigraph obtained from ƒi by doubling all edges of ƒi which arenot traversed by Wi . Specifically, V.ƒ0i/D V.ƒi/ and for u; v 2 V.ƒ0i/: there is no


190 Brandon Seward

edge between u and v in ƒ0i if .u; v/ 62 E.ƒi/; there is a single edge between u and vin ƒ0i if .u; v/ 2 E.ƒi/ and Wi traverses .u; v/; there are two edges between u and vin ƒ0i if .u; v/ 2 E.ƒi/ and Wi does not traverse .u; v/.

We claim that p is the unique vertex of odd degree in ƒ0i . If u 2 V.ƒ0i/ is not visitedby the path Wi , then u is connected to each of its neighbors via 2 edges in ƒ0i andthus u has even degree in ƒ0i . If u ¤ p and u is visited by the path Wi , then Wi

traverses precisely two distinct edges (in ƒi ) adjacent to u. All edges adjacent to u

and not traversed by Wi were doubled in passing to ƒ0i . Thus u has even degreein ƒ0i . Now it only remains to show that p has odd degree in ƒ0i . Since only one edgeadjacent to p was traversed by Wi , all edges adjacent to p but one were doubled inpassing to ƒ0i . Thus p has odd degree in ƒ0i , completing the proof of the claim.

We claim that the multigraph ƒ01[ƒ0

2satisfies the conditions (i), (ii) and (iii) of the

Erdos–Grünwald–Weiszfeld Theorem. Notice that p is the only vertex contained inboth ƒ0

1and ƒ0

2. Since p has odd degree in each ƒ0i , it has even degree in ƒ0

1[ƒ0

2.

So clauses (i) and (ii) are clearly satisfied. We consider (iii). In our argument we willmake use of the fact that if a finite multigraph has at most one vertex of odd degree,then it has no vertices of odd degree at all. This follows from the fact that the sumsof the degrees of the vertices is always twice the number of edges. Consider a finiteeven subgraph ˆ of ƒ0

1[ƒ0

2. For i D 1; 2 set î D ˆ\ƒ

0i . Since ˆ is even, î

has at most one vertex of odd degree, namely p . Since î is finite, it has no verticesof odd degree. Therefore î is a finite even subgraph of ƒ0i . Thus every vertex ofƒ0i�E.î/ has the same degree modulo 2 as it has in ƒ0i . Thus p is the unique vertexof odd degree in ƒ0i �E.î/. So any finite connected component of ƒ0i �E.î/ canhave at most one vertex of odd degree and therefore cannot have any vertices of odddegree. So p must lie in an infinite connected component of ƒ0i �E.î/. However, �has two ends and from how we constructed ƒ0

1and ƒ0

2one can see that each ƒ0i

must be one ended. So p is contained in the unique infinite connected componentsof ƒ0

1� E.ˆ1/ and ƒ0

2� E.ˆ2/. Thus ƒ0

1[ƒ0

2� E.ˆ/ has precisely one infinite

connected component. By the Erdos–Grünwald–Weiszfeld Theorem we get an Eulerianpath

P W Z! V.ƒ01/[V.ƒ02/D V.�/:

Clearly P is a path on � which visits every vertex at least once and traverses everyedge at most twice.

Regardless of whether � has one end or two, we have a path P W Z! V.�/ whichvisits every vertex at least once and traverses every edge at most twice. Now we will usethis path P to show that � is bilipschitz equivalent to a graph admitting a bi-infiniteHamiltonian path. In order to do this, we will need to apply Hall’s Marriage Theorem.



Recall that a graph ƒ is bipartite if there is a partition fV1;V2g of V.ƒ/ such thatevery edge of ƒ joins a vertex of V1 to a vertex of V2 . Hall’s Marriage Theorem statesthat if ƒ is a locally finite bipartite graph with bipartition fV1;V2g of V.ƒ/ and ifjNƒ

1.T /�T j � jT j for all T � V1 , then there is an injection �W V1! V2 with the

property that v and �.v/ are joined by an edge for every v 2 V1 (the statement forfinite graphs is [2, Section III.3, Theorem 7]; the theorem extends to infinite graphs bya standard compactness argument).

Let D be a positive integer satisfying deg�.v/�D for all v 2V.�/. Set M DDC1.Let ƒ be the bipartite graph with vertex set M Z[V.�/ and edge relation given by

.k �M; v/ 2 E.ƒ/ ” 9 0� i <M; P .k �M C i/D v:

Fix a finite T � M Z and set @T D Nƒ1.T / � T � V.�/. We wish to show that

j@T j � jT j. Notice that

@T D P .T /[P .T C 1/[ � � � [P .T CM � 1/:

Let ˆT � � be the graph with vertex set @T and edge relation

f.P .t C i/;P .t C i C 1// j t 2 T; 0� i �M � 2g:

Since P traverses each edge at most twice, ˆT has at least 12jT j.M � 1/ edges.

Since P is a path in �, every vertex of ˆT has degree at most DDM �1. Therefore

.M � 1/jV.ˆT /j �X

v2V .ˆT /

degˆT.v/D 2 � jE.ˆT /j � jT j.M � 1/

and hencej@T j D jV.ˆT /j � jT j:

Thus the condition for Hall’s Marriage Theorem is satisfied, and therefore there is afunction �W M Z! Œ0;M � 1� satisfying .k;P .kC�.k/// 2 E.ƒ/ for all k 2M Zand P .k1C�.k1//¤ P .k2C�.k2// for all k1 ¤ k2 2M Z.

Set AD fP .kC�.k// j k 2M Zg �V.�/. For v 2V.�/�A, pick any nv 2Z withP .nv/D v . Set

S D fkC�.k/ j k 2M Zg[ fnv j v 2 V.�/�Ag:

Clearly the restriction of P to S is a bijection between S and V.�/. Let W Z!S bean order preserving bijection, and define QW Z!V.�/ by Q.z/DP . .z//. Then Q

is a bijection. By the definition of S , it is clear that consecutive numbers in S areseparated by a distance of at most 2M � 1 and hence .k C 1/� .k/ � 2M � 1

for all k 2 Z. Since P is a path in �, the distance between Q.k C 1/ and Q.k/ is


192 Brandon Seward

at most 2M � 1. Now let � 0 be the graph with vertex set V.�/ and let there be anedge between v0 and v1 if and only if there is a path in � of length at most 2M � 1

joining v0 to v1 . Then � and � 0 are bilipschitz equivalent and Q is a Hamiltonianpath on � 0 .

Corollary 3.4 Let � be a connected infinite graph whose vertices have uniformlybounded degree. Then � admits a transitive translation-like action by Z if and onlyif � has at most two ends.

Proof This follows immediately from the previous theorem and Lemma 3.1.

Among the graphs which are infinite, connected, locally finite and have uniformlybounded degree, the above theorem completely classifies which of these graphs arebilipschitz equivalent to a graph admitting a bi-infinite Hamiltonian path. It wouldbe interesting if nice characterizations of this property could be found among graphswhich are not locally finite, or which are locally finite but whose vertices do not haveuniformly bounded degree.

Problem 3.5 Find necessary and sufficient conditions for a graph to be bilipschitzequivalent to a graph admitting a bi-infinite Hamiltonian path.

4 Geometric Burnside’s Problem

The theorem of the previous section allows us to easily resolve the Geometric Burnside’sProblem.

Theorem 4.1 (Geometric Burnside’s Problem) Every finitely generated infinite groupadmits a translation-like action by Z.

Proof Let G be a finitely generated infinite group. If G has finitely many ends,then G has at most two ends [4, Section I.8, Theorem 8.32]. Now consider any Cayleygraph of G and apply Corollary 3.4 to get a translation-like action of Z on G (noticethat this action is in fact transitive). Now suppose that G has infinitely many ends.Then by Stallings’ Theorem [4, Section I.8, Theorem 8.32, clause (5)], G can beexpressed as an amalgamated product A�C B or HNN extension A�C with C finite,ŒA W C �� 3, and ŒB W C �� 2. If G DA�C B then ab 2G has infinite order whenevera 2A�C and b 2 B �C. If G DA�C D hA[ftg j 8c 2 C; tct�1 D �.c/i, where�W C ! A is an injective homomorphism, then t has infinite order. Thus in eithercase, there is a subgroup H �G with H ŠZ. Then the natural right action of ZŠH

on G is a translation-like action (see Proposition 1.2).



Between the theorem above and Whyte’s result in [14], we have that Burnside’s Problemand the von Neumann Conjecture both hold true in the geometric setting where subgroupcontainment is replaced by the existence of a translation-like action. It is interestingto wonder how much further these types of results can be taken. Two very specificquestions in this direction are the following.

Problem 4.2 (Whyte [14]) Is the Geometric Gersten Conjecture true? (See clause(3 0 ) in Section 1 above).

Problem 4.3 Does a finitely generated group have exponential growth if and only ifit admits a translation-like action by a free semigroup on two generators?

The algebraic version of the above problem would be: Does a finitely generated grouphave exponential growth if and only if it contains a free subsemigroup on two generators?Chou [5] proved that every elementary amenable group of exponential growth containsa free subsemigroup on two generators. However, the algebraic version is false ingeneral. Grigorchuk has pointed out that the wreath product of the Grigorchuk groupwith the cyclic group of order 2 is a torsion group of exponential growth. Since it istorsion, it cannot contain a free subsemigroup. The geometric (translation-like action)version may hold true. By Chou’s result, it holds for elementary amenable groups, andby the Geometric von Neumann Conjecture it holds for nonamenble groups.

A more general, open-ended question is the following.

Problem 4.4 What other algebraic/geometric properties can be reformulated or char-acterized in terms of translation-like actions?

Before ending this section we present two corollaries treating Hamiltonian paths onCayley graphs and translation-like actions of Z which are transitive.

Corollary 4.5 A finitely generated infinite group has finitely many ends if and only ifit admits a translation-like action by Z which is transitive.

Proof If G has finitely many ends then the proof of the previous theorem shows thatthere is a translation-like action of Z on G which is transitive. Now suppose that thereis a translation-like action of Z on G which is transitive. If Cay.G/ is any Cayleygraph of G then the action of Z on G gives rise to a transitive translation-like actionof Z on Cay.G/. By Corollary 3.4, Cay.G/ has at most two ends and thus G has atmost two ends.

Notice that the following corollary no longer requires the group to be infinite.


194 Brandon Seward

Corollary 4.6 A finitely generated group has finitely many ends if and only if it has aCayley graph admitting a Hamiltonian path.

Proof For infinite groups, this follows from the above corollary and Corollary 3.2.Finally, every finite group has 0 ends (in particular, has finitely many ends), andthe complete graph on jGj vertices is a Cayley graph of G which clearly admits aHamiltonian path.

Relating to the above corollary, we point out that Igor Pak and Radoš Radoicic provedin [10] that every finite group G has a generating set S with jS j � log2 jGj suchthat the Cayley graph Cay.GIS/ admits a Hamiltonian path. This ties in with thewell-known Lovász Conjecture: Every finite, connected, vertex-transitive graph admitsa Hamiltonian path. Clearly this conjecture applies to finite Cayley graphs. Despite thePak–Radoicic result, the special case of the Lovász Conjecture for Cayley graphs is farfrom settled.

Problem 4.7 (Special case of the Lovász Conjecture) Does every Cayley graph ofevery finite group admit a Hamiltonian path?

Problem 4.8 Does every Cayley graph of every finitely generated group with finitelymany ends admit a Hamiltonian path?

5 Regular spanning trees of Cayley graphs

In this section we will prove that every finitely generated infinite group has a Cayleygraph containing a regular spanning tree. To do this we will strengthen Whyte’s resulton the Geometric von Neumann Conjecture.

We first present two simple properties of translation-like actions. These properties werementioned in [14] without proof. For convenience to the reader we include a proofhere.

Lemma 5.1 (Whyte [14]) Let H be a finitely generated group, let .X; d/ be a metricspace and suppose H acts on X and the action is translation-like. Then there is aconstant C such that for all x 2X the map

h 2H 7! x � h 2X

is a C–Lipschitz injection.



Proof Let U be a finite generating set for H and let � be the corresponding left-invariant word length metric. For each u2U [U�1 let cu� 1 satisfy d.x;x�u/� cu

for all x 2X. Set C Dmaxfcu j u 2U [U�1g and fix x 2X. For h; k 2H we have

d.x � k;x � h/D d..x � h/� h�1k;x � h/� C � �.h�1k; 1H /D C � �.k; h/:

So the map is C–Lipschitz as claimed. Also, the map is injective as translation-likeactions are free.

Corollary 5.2 (Whyte [14]) Let H and G be finitely generated groups, let Cay.H /

be a Cayley graph of H and suppose that H acts on G and the action is translation-like.Then there is a Cayley graph Cay.G/ of G so that for all g 2G and h0; h1 2H with.h0; h1/ 2 E.Cay.H //, .g�h0;g�h1/ 2 E.Cay.G//. In particular, there is a spanningsubgraph ˆ of Cay.G/ such that the connected components of ˆ are precisely theorbits of the H action and every connected component of ˆ is graph isomorphicto Cay.H /.

Proof Let V D V �1 be a finite generating set for G, and let d be the correspondingleft-invariant word length metric. Let Cay.H / be a Cayley graph of H , say Cay.H /D

Cay.H IU /, where U DU�1 is a finite generating set for H . Let � be the left-invariantword length metric on H corresponding to U . Let C be as in the previous lemma. Bypicking a larger C if necessary, we may assume C 2N . Set W D V [V 2[� � �[V C.Then W is a generating set for G and for all g 2 G and u 2 U � H we haveg �u 2 gW . Consider the Cayley graph Cay.GIW /. Fix g 2G and h0; h1 2H with.h0; h1/ 2 E.Cay.H IU //. Then there is u 2 U with h1 D h0u. Therefore

g � h1 D .g � h0/�u 2 .g � h0/W;

so .g � h1;g � h0/ 2 E.Cay.GIW //. Thus the first claim holds. Now let ˆ be thegraph with vertex set G and edge set f.g;g � u/ j g 2 G; u 2 U g. Then ˆ is aspanning subgraph of Cay.GIW /, the connected components of ˆ are the orbits ofthe H action, and every connected component of ˆ is graph isomorphic to Cay.H IU /(since the action of H is free).

Let ƒ be a tree with a distinguished root vertex �. Define a partial ordering, denoted �,on V.ƒ/ by declaring u � v if and only if the unique shortest path from � to v

traverses u.

Definition 5.3 We call a finite set P �V.ƒ/ a perimeter if P ¤f�g and the followingconditions are met:

(i) There is a constant R such that if d.u; �/�R then there is p 2 P with p � u.

(ii) If p;p0 2 P and p � p0 , then p D p0 .


196 Brandon Seward

The smallest such R satisfying (i) is the radius of P .

Perimeters will soon be used to help us construct quasi-isometries between trees. Theidea is that if ƒ and T are trees with distinguished roots, F W V.T /! V.ƒ/ is apartially defined function, and F.t/ is defined, then we want to extend F so that thevertices of T adjacent to t are mapped, not necessarily injectively, to a perimeterof F.t/. The following is a technical lemma needed for the proposition following itwhere this construction is carried out. From the viewpoint of the construction justdiscussed, in the lemma below r essentially represents the number of vertices of T

adjacent to t , and d.p/ tells us how many vertices of T adjacent to t should bemapped to p .

Lemma 5.4 Let ƒ be a tree with distinguished root vertex �, let � be defined asabove, and let r � 3. If 2� deg.�/� r � 1 and all the other vertices of ƒ have degreeat least 3, then there is a perimeter P of radius at most r and a function d W P !NC

satisfying d.p/� deg.p/� 1 for all p 2 P andXp2P

d.p/D r:

Proof The proof is by induction on r . If r D 3, then deg.�/D 2. Let P be the set ofvertices adjacent to �, say P D fa; bg. Define d.a/D 2 and d.b/D 1. Then P and d

have the desired properties. Now suppose that r � 4 and that the claim is true for allr 0 < r . We have deg.�/� r �1, so by the Euclidean algorithm there are q; s 2N withs < deg.�/ and

r D q � deg.�/C s D q.deg.�/� s/C .qC 1/s:

Clearly q � rdeg.�/ �

r2

. Since r � 4, we have q; qC 1< r .

Divide the vertices adjacent to � into disjoint sets A and B with jAj D deg.�/� s > 0

and jBj D s (note that we can have sD 0 in which case BD¿). Fix a vertex a2A. Ifq� deg.a/�1, then set PaDfag and da.a/D q . If q> deg.a/�1, then let ƒa be thegraph with vertex set fu2V.ƒ/ ja�ug and define there to be an edge between u and vif and only if they are joined by an edge in ƒ. Then ƒa is a tree, and we declare a tobe its root vertex. Clearly degƒa

.a/D degƒ.a/� 1 and degƒa.u/D degƒ.u/ for all

a¤ u 2 V.ƒa/. By the inductive hypothesis, there is a perimeter Pa of ƒa of radiusat most q and a function daW Pa!NC with da.p/� degƒa

.p/� 1D degƒ.p/� 1

for all p 2 Pa and Xp2Pa

da.p/D q:



Carry out the same construction for every element of B , but with q replaced by qC 1.Set

P D[

c2A[B

Pc ; d D[

c2A[B

dc :

Then P is a perimeter for ƒ, d.p/� degƒ.p/� 1 for all p 2 P, andXp2P

d.p/DXa2A

Xp2Pa

da.p/CXb2B

Xp2Pb

db.p/D qjAjC .qC 1/jBj D r:

Also, the radius of P is at most

qC 2� r2C 2� r:

Theorem 5.5 Let ƒ1 and ƒ2 be two trees. If every vertex of ƒ1 and ƒ2 has degreeat least three and if the vertices of ƒ1 and ƒ2 have uniformly bounded degree, then ƒ1

and ƒ2 are bilipschitz equivalent.

Proof Let r > 2 be such that every vertex of ƒ1 and ƒ2 has degree at most r . Let T

be the regular tree of degree r C 1. It suffices to show that ƒ1 and ƒ2 are bothbilipschitz equivalent to T . So dropping subscripts, let ƒ be a tree such that everyvertex has degree at least 3 and at most r . We will show that ƒ and T are bilipschitzequivalent.

We will first define a quasi-isometry F W V.T / ! V.ƒ/. Recall that a functionf W .X; d/! .Y; �/ between metric spaces .X; d/ and .Y; �/ is called a quasi-isometryif there are constants K and C such that

1K

d.x1;x2/�C � �.f .x1/; f .x2//�Kd.x1;x2/CC

and such that for every y 2 Y there is x 2X with �.y; f .x// <C. Also recall that X

and Y are quasi-isometric if there is a quasi-isometry f W X ! Y . We will view V.T /and V.ƒ/ as metric spaces by using the path length metrics described in Section 2.Let d and � denote the path length metrics on V.T / and V.ƒ/, respectively. Fix rootvertices t0 2 V.T / and �0 2 V.ƒ/. Define a partial ordering, denoted �, on V.T /by declaring u � v if and only if the unique shortest path from t0 to v traverses u.Similarly define a partial ordering, also denoted �, on V.ƒ/. Define

S.T I n/D fu 2 V.T / j d.t0;u/D ng; B.T I n/D fu 2 V.T / j d.t0;u/� ng:

Clearly V.T / is the union of the S.T I n/. We will inductively define a quasi-isometryF W V.T /! V.ƒ/ so that the following properties hold:


198 Brandon Seward

(i) F.t0/D �0 .

(ii) If F.u/D F.v/, then d.u; v/� 2 and d.t0;u/D d.t0; v/.

(iii) If d.u; v/D 1 then �.F.u/;F.v//� r C 1.

(iv) If u� v then F.u/� F.v/.

(v) If F.u/� F.v/ then there is w 2 F�1.u/ with w � v .

(vi) For y 2 V.ƒ/, jF�1.y/j � degƒ.y/� 1.

(vii) If u; v 2 S.T I n/ and F.u/� F.v/, then F.u/D F.v/.

(viii) For every n� 1 F.S.T I n// is a perimeter.

Set F.t0/ D �0 so that (i) is satisfied. By Lemma 5.4, there is a perimeter P of ƒof radius at most r C 1 and a function d W P ! NC with d.p/ � degƒ.p/� 1 forall p 2 P and

Pp2P d.p/ D r C 1. Since jS.T I 1/j D r C 1, there is a surjection

F W S.T I 1/!P such that jF�1.p/jDd.p/. The definition of F on B.T I 1/ satisfiesclauses (i) through (viii).

Now suppose that F has been defined on B.T I n� 1/ and does not violate any of theclauses (i) through (viii). We will define F on S.T I n/. Notice that (v) and (ii) tellus that if F.u/ � F.v/ then d.t0;u/ � d.t0; v/. Therefore by (vii) we have that ifu 2 S.T I n� 1/ then there is nothing strictly �–larger than F.u/ in F.B.T I n� 1//.This together with (viii) imply that the �–maximal elements of F.B.T I n� 1// areprecisely the F images of elements of S.T I n� 1/. Let y1;y2; : : : ;ym be the �–maximal elements of F.B.T I n�1//, and set Ui DF�1.yi/\S.T I n�1/¤¿. Thenthe Ui form a partition of S.T I n� 1/. To define F on S.T I n/, it suffices to fix an i

and define F on the elements of S.T I n/ which are adjacent to an element of Ui . Thisdefinition is provided in the following paragraph. To simplify notation, in the paragraphbelow we work with a single yi and Ui but omit the subscript i .

By (vi), we have jU j � degƒ.y/ � 1. Fix any function hW U ! NC satisfyingPv2U h.v/D degƒ.y/� 1. Partition the set of vertices which are adjacent to y and

�–larger than y into disjoint sets fAv j v 2U g such that jAvj D h.v/. For each v 2U

we will define a graph ƒv . We first describe the vertex set. If jAvj � 2, then welet ƒv be the graph with vertex set the union of y with Av and with all of the verticesof ƒ which are �–larger than some element of Av . If jAvj D 1 then we let ƒv havethe same vertices as described in the previous sentence, except that we exclude y

from the vertex set. In either case, we let there be an edge between w1; w2 2 V.ƒv/if and only if they are joined by an edge in ƒ. Notice that ƒv is a subgraph of ƒ,degƒv

.y/DjAvj�degƒ.y/�1 (if y 2V.ƒv/), degƒv.a/Ddegƒ.a/�1 (if AvDfag)

and degƒv.w/D degƒ.w/ for all other vertices w 2V.ƒv/. We declare the root vertex



of ƒv to be y if y 2 V.ƒv/ and otherwise to be a if Av D fag. By Lemma 5.4,there is a perimeter Pv of ƒv of radius at most r and a function dvW Pv!NC withdv.p/� degƒv

.p/� 1D degƒ.p/� 1 for all p 2 Pv andXp2Pv

dv.p/D r:

Thus, if S.v/ denotes the set of vertices of T adjacent to v and �–larger than v , thenjS.v/j D r and there is a surjection F W S.v/! Pv with jF�1.p/j D dv.p/ for allp 2 Pv . This defines F on S.v/ for each v 2 U . Thus we have defined F on allelements of S.T I n/ which are adjacent to some element of U . By varying the set U ,we get a definition of F on S.T I n/ and thus on B.T I n/. It is easy to check that F

still satisfies clauses (i) through (viii). By induction, this completes the definition of F.

We now verify, relying only on clauses (i) through (viii), that F is a quasi-isometry.Clearly clause (iii) implies that �.F.u/;F.v//� .rC1/d.u; v/ for all u; v2V.T /. Wenow check by cases that for all u; v 2V.T / we have d.u; v/� r �2� �.F.u/;F.v//.Fix u; v 2V.T / and let w 2V.T / be the �–largest vertex with w � u; v . Notice thatclauses (ii) and (iv) imply that if p � q then d.p; q/� �.F.p/;F.q//.

Case 1: d.w;u/� 1 or d.w; v/� 1 Without loss of generality, suppose d.u; w/�1.Then �.F.u/;F.w//� r C 1 by (iii) and d.w; v/� �.F.w/;F.v// since w � v . So

d.u; v/� r � 2D d.u; w/C d.w; v/� r � 2

� �.F.w/;F.v//� �.F.w/;F.u//

� �.F.u/;F.v//:

Case 2: d.w;u/;d.w; v/� 2 Let wu and wv be the unique vertices of T withd.w;wu/D d.w;wv/D 2, w �wu � u and w �wv � v . Notice that d.wu; wv/D 4.By (v) and (ii), F.wu/ and F.wv/ are not �–comparable. Thus �.F.wu/;F.wv//�2,and since wu � u and wv � v , we have that d.wu;u/ � �.F.wu/;F.u// andd.wv; v/� �.F.wv/;F.v//. Thus

d.u; v/� 2D d.u; wu/C d.v; wv/C d.wu; wv/� 2

� �.F.u/;F.wu//C �.F.v/;F.wv//C �.F.wu/;F.wv//

D �.F.u/;F.v//:

To show that F is a quasi-isometry, all that remains is to show that for all y 2 V.ƒ/there is u2V.T / with �.y;F.u//� rC1. So fix y 2V.ƒ/ and towards a contradictionsuppose that every image point of F is at least a distance of rC2 from y . First supposethat there is u 2 V.T / with y � F.u/. Let P be a path in T from t0 to u. Let n beleast such that y 6�F.P .n// and y �F.P .nC1//. Since F.P .n// and F.P .nC1//


200 Brandon Seward

are distance at least r C 2 from y , we have �.F.P .n//;F.P .nC 1/// � 2r C 4,contradicting (iii). So there are no u 2 V.T / with y � F.u/. Since each F.S.T I n//

is a perimeter, it follows that for each n 2N there is un 2 S.T I n/ with F.un/� y .However, there are only finitely many w 2 V.ƒ/ with w � y , so this violates clause(vi). We conclude that every vertex of ƒ is within a distance of r C 1 of some imagepoint of F. Thus F is a quasi-isometry, and so ƒ is quasi-isometric to T .

Recall that a graph � is nonamenable if there exist C > 1 and k 2 N such thatjN�

k.S/j � C � jS j for all finite S � V.�/, where N�

k.S/ is the union of S with all

vertices of � within distance k of S . When � is the Cayley graph of a group, thisnotion coincides with the standard notion of amenability of groups. It is known thatif � is a tree and every vertex of � has degree at least 3, then � is nonamenable(one can use C D 2 and k D 1). By [8, Section IV.B, Complement 46], any twononamenable quasi-isometric graphs are bilipschitz equivalent. Since ƒ and T arenonamenable and quasi-isometric, they are bilipschitz equivalent.

In the following theorem we strengthen Whyte’s result on the Geometric von NeumannConjecture. In the remainder of this section, we let Fk denote the nonabelian freegroup of rank k for k > 1.

Theorem 5.6 (Geometric von Neumann Conjecture) Let G be a finitely generatedgroup. The following are equivalent for every integer k � 2:

(i) G is nonamenable.

(ii) G admits a translation-like action by Fk .

(iii) G admits a transitive translation-like action by Fk .

Proof In [14], Whyte proved the equivalence of (i) and (ii). Clearly (iii) implies (ii),so we only must show that (ii) implies (iii). Let Cay.Fk/ be the canonical Cayleygraph of Fk . So Cay.Fk/ is a regular tree of degree 2k . By assumption, there is atranslation-like action � of Fk on G. By Corollary 5.2, there is a Cayley graph Cay.G/of G having a spanning subgraph ˆ in which the connected components of ˆ are theorbits of the Fk action and every connected component of ˆ is graph isomorphic toCay.Fk/, the regular tree of degree 2k .

Let ‰ be the quotient graph of Cay.G/ obtained by identifying points which lie in acommon Fk orbit. Specifically, the vertex set of ‰ is fg �Fk j g 2Gg and there is anedge between g �Fk and h �Fk if and only if there are t0; t1 2 Fk such that g � t0and h� t1 are joined by an edge in Cay.G/. Since Cay.G/ is connected, so is ‰ . Weconstruct a spanning tree ‰0 of ‰ as follows. Set 0D 1G �Fk 2V.‰/, and let d be



the path length metric on ‰ . For each 0¤ v 2V.‰/, pick any f .v/2V.‰/ with theproperty that d.v; f .v//D 1 and d. 0; f .v//D d. 0; v/�1. We set V.‰0/DV.‰/and

E.‰0/D f.v; f .v// j 0 ¤ v 2 V.‰/g[ f.f .v/; v/ j 0 ¤ v 2 V.‰/g:

A simple inductive argument on the magnitude of d. 0; v/ shows that ‰0 is con-nected. Also, if u and v are joined by an edge in ‰0 then d. 0;u/¤ d. 0; v/ andd. 0;u/ < d. 0; v/ implies uD f .v/. Thus ‰0 does not contain any cycles as cyclescannot contain a vertex of maximal distance to 0 . So ‰0 is a spanning tree of ‰ .

Recall that ˆ is a subgraph of Cay.G/ in which the connected components are regulartrees of degree 2k . Also notice that the vertices of ‰ and ‰0 are precisely the connectedcomponents of ˆ (since both are just the orbits of the Fk action). We will use ‰0

to enlarge ˆ to a spanning tree ˆ0 of Cay.G/. Let e 2 E.‰0/ be an edge in ‰0 , saybetween g1 � Fk and g2 � Fk . Then by definition, there are t1; t2 2 Fk such thatg1 � t1 and g2 � t2 are joined by an edge in Cay.G/. Let h.e/ be any such edge.g1 � t1;g2 � t2/ 2 E.Cay.G//. So hW E.‰0/! E.Cay.G//. Now let ˆ0 be the graphwith vertex set G and edge set h.E.‰0//[ E.ˆ/. Since the connected componentsof ˆ are trees, the vertices of ‰0 are the connected components of ˆ, and ‰0 is atree, it is easy to see that ˆ0 is also a tree. So ˆ0 is a spanning tree of Cay.G/. Eachvertex of ˆ0 has degree at least 2k , and the degrees of the vertices of ˆ0 are uniformlybounded since ˆ�ˆ0 � Cay.G/. By Theorem 5.5, ˆ0 (with its path length metric)is bilipschitz equivalent to the 2k regular tree Cay.Fk/. Let P W Fk ! V.ˆ0/ be thisbilipschitz equivalence. Since ˆ0 is a subgraph of Cay.G/, it follows that the mapP W Fk !G is Lipschitz when G D V.ˆ0/ is equipped with the path length metric ofCay.G/. Now we define a transitive translation-like action ı of Fk on G as follows.For t 2 Fk and g 2G, we define

g ı t D P .P�1.g/ � t/;

where � denotes the group operation of Fk . This is clearly a transitive translation-likeaction since P W Fk !G is a Lipschitz bijection (where G has the path length metriccoming from Cay.G/).

We point out that in the previous proof, instead of using Whyte’s result we could havealso used a similar result by Benjamini and Schramm [1, Remark 1.9].

Combining the previous theorem with Corollary 4.5 gives the following.

Corollary 5.7 Every finitely generated infinite group admits a transitive translation-like action by some (possibly cyclic) free group.


202 Brandon Seward

Proof By Corollary 4.5, G admits a transitive translation-like action by Z (which isa cyclic free group) if G has finitely many ends. So now suppose that G has infinitelymany ends. We will show that G contains a nonabelian free group of rank 2 and is thusnonamenable. By Stallings’ Theorem [4, Section I.8, Theorem 8.32, clause (5)], G

can be expressed as an amalgamated product A�C B or HNN extension A�C with C

finite, ŒA WC �� 3 and ŒB WC �� 2. We now show by cases that G contains a nonabelianfree group of rank 2.

Case 1: C is trivial and G is of the form A �C B DA �B By [12, Proposition 4],the kernel of the homomorphism A � B ! A � B is a free group with free basisfa�1b�1ab j 1A ¤ a 2A; 1B ¤ b 2 Bg. So we only need to show that this free basiscontains more than one element. Since jAj � 3, there are nonidentity a1¤ a2 2A. Fixany nonidentity b 2B . Since G DA�B , it is clear that a�1

1b�1a1b and a�1

2b�1a2b

are distinct. Thus the free basis of the kernel contains at least two elements andtherefore G contains a nonabelian free group of rank 2.

Case 2: C is nontrivial and G is of the form A �C B We use [4, Section III.�.6,Lemma 6.4]. This lemma states, in particular, that if a1 2A, a2; a3; : : : ; an 2A�C,b1; : : : ; bn�1 2B�C and bn 2B , then a1b1a2b2 � � � anbn is not the identity elementin G DA�C B . Pick any b1; b2 2B�C, and pick any a1 2A�C. Since ŒA WC �� 3,we can also pick a2 2 A� C with a2a1 62 C. If b2b1 2 C, then pick a3 2 A� C

with a3b2b1a1 62 C. If b2b1 62 C, then pick any a3 2 A� C. Set u D a1b1a2 andv D b1a1b2a3b2 . Then u and v generate a nonabelian free group of rank 2.

Case 3: G is of the form A�C D hA; tjt�1ct D �.c/g, where ŒAW C �� 3 and

�W C !A is an injective homomorphism Let C 0 D �.C /. We again use [4, Sec-tion III.�.6, Lemma 6.4]. This lemma states, in particular, that if a1; a2; : : : ; an�1 2

A� .C [C 0/ and m1;m2; : : : ;mn 2 Z� f0g then tm1a1tm2a2 � � � tmn�1an�1tmn is

not the identity element in G DA�C . Since C is finite, C 0 cannot properly contain C

and therefore C 0 cannot contain any coset aC with a 62 C. Therefore, we have thatthere is a 2 A � C [ C 0 . Set u D tat and v D t2at2 . Then u and v generate anonabelian free group of rank 2.

Thus G contains a nonabelian free group of rank 2. The group G is therefore nona-menable (this is a well known consequence of containing a nonabelian free group butalso follows directly from the previous theorem), and so by the previous theorem thereis a transitive translation-like action of the nonabelian free group of rank 2 on G.

The previous corollary has an interesting reinterpretation in terms of Cayley graphs.



Corollary 5.8 If G is a finitely generated infinite group then G has a Cayley graphadmitting a regular spanning tree. In fact, for every integer k > 2 the following hold:

(i) G has finitely many ends if and only if G has a Cayley graph admitting aHamiltonian path (ie a regular spanning tree of degree 2).

(ii) G is nonamenable if and only if G has a Cayley graph admitting a regularspanning tree of degree k .

Proof For the first statement, apply the previous corollary and Corollary 5.2 andnotice that the graph ˆ referred to in Corollary 5.2 has only one connected componentsince the action is transitive. Clause (i) is Corollary 4.6.

Now fix k > 2. If G is nonamenable, then by clause (iii) of Theorem 5.6 andCorollary 5.2 G has a Cayley graph Cay.G/ admitting a regular spanning tree ˆof degree 2k . Say Cay.G/ D Cay.GIV /, where V is a finite generating set forG, and let d be the corresponding left-invariant word length metric. Let ƒ be theregular tree of degree k . By Theorem 5.5, ˆ and ƒ are bilipschitz equivalent, say viaf W V.ƒ/!V.ˆ/. Since ˆ is a subgraph of Cay.G/, it follows that there is a constantc 2 N so that d.f .t0/; f .t1// � c whenever t0; t1 2 V.ƒ/ are adjacent in ƒ. SetW DV [V 2[� � �[V c . Then W is a finite generating set for G. Consider Cay.GIW /.Clearly we have that .f .t0/; f .t1// 2 E.Cay.GIW // whenever .t0; t1/ 2 E.ƒ/. Thus,if we define ˆ0 to be the graph with vertex set G and edge set ff .e/ j e 2 E.ƒ/g,then ˆ0 is a subgraph of Cay.GIW /. Since f is bijective with V.ˆ/ D G, ˆ0 isgraph isomorphic to ƒ and is a regular spanning tree of Cay.GIW / of degree k .

Now suppose that G has a Cayley graph, Cay.G/, admitting a regular spanning tree ˆof degree k . Then ˆ is nonamenable in the sense that there are C > 1 and m 2N sothat jNˆ

m .S/j � C � jS j for all finite S � V.ˆ/. Since ˆ is a spanning subgraph of �,it follows that jN�

m.S/j � C � jS j for all finite S � V.�/ (since N�m.S/�Nˆ

m .S/).Thus, by Følner’s characterization of amenability, G is not amenable.

The fact that every finitely generated infinite group has some Cayley graph admitting aregular spanning tree is a rather peculiar fact, for as the following proposition shows,not all Cayley graphs have this property.

Proposition 5.9 There is a finitely generated infinite group G and a Cayley graphCay.G/ of G which does not have any regular spanning tree.

Proof Let G D Z �Z3 . Say Z D hti and Z3 D hui. Let S D ft;ug. Then S isa finite generating set for G. We claim that Cay.GIS/ does not contain any regularspanning trees.


204 Brandon Seward

Due to the structure of free products, it is not difficult to draw the Cayley graphCay.GIS/. This graph can be obtained by starting with a regular tree of degree 6,replacing each of the vertices with a triangle, and evenly dividing the 6 edges adjacentto the original vertex between the 3 vertices of the new triangle in its place. Thus everyvertex of Cay.GIS/ lies in a triangle (corresponding to repeated multiplication on theright by u), and is adjacent to two other triangles (by multiplying on the right by t

and t�1 ). If the triangles in Cay.GIS/ are collapsed, then a 6–regular tree is obtained.A finite portion of Cay.GIS/ is displayed in Figure 1.

Figure 1: A finite portion of the Cayley graph Cay.GIS/

Now let ƒ be a spanning tree of Cay.GIS/. It suffices to show that ƒ is not regular.Define

Et D f.g;gt/ j g 2Gg[ f.g;gt�1/ j g 2Gg;

Eu D f.g;gu/ j g 2Gg[ f.g;gu�1/ j g 2Gg:

Notice that E.Cay.GIS// is the disjoint union Et [Eu . From the description inthe previous paragraph, it is clear that E.ƒ/ must contain Et as otherwise ƒ wouldeither not be spanning or not be connected. Notice that every vertex of Cay.GIS/is adjacent to precisely two edges from Et . Now consider the three edges .1G ;u/,.u;u2/D .u;u�1/ and .u�1; 1G/. In order for ƒ to be spanning, connected and a tree,E.ƒ/ must contain precisely two of these three edges. Thus for some i D�1; 0; 1 wehave .ui ;uiC1/; .ui ;ui�1/ 2 E.ƒ/ and .uiC1;ui�1/ 62 E.ƒ/. So degƒ.u

i/D 4 anddegƒ.u

iC1/D degƒ.ui�1/D 3. Thus ƒ is not regular. We conclude that Cay.GIS/

does not contain any regular spanning trees.

Looking back at the clauses of Corollary 5.8, we see that the above proposition onlyshows that “a Cayley graph” cannot be replaced by “every Cayley graph” in clause



(ii). However, it is unclear if clause (i) remains true if “a Cayley graph” is replaced by“every Cayley graph.”

Problem 5.10 Does every infinite Cayley graph with finitely many ends admit aregular spanning tree?

Problem 5.11 If G is a finitely generated nonamenable group with finitely many ends,then does every Cayley graph of G admit a regular spanning tree of degree strictlygreater than two?

Corollary 5.8 says that if G is finitely generated and nonamenable, then for everyk � 3 there is a Lipschitz bijection from the k –regular tree onto G. In general thesemaps cannot be improved to being bilipschitz since some nonamenable groups havefinitely many ends and the number of ends is invariant under bilipschitz equivalences.However, it is unclear when we can get a bilipschitz map at the expense of losingbijectivity. By a result of Benjamini and Schramm in [1], every nonamenable groupcontains a bilipschitz image of a regular tree of degree 3. However, it is not clear ifany amenable group contains a bilipschitz image of a regular tree of degree 3.

Problem 5.12 Which finitely generated groups contain a bilipschitz image of the3–regular tree?

Notice that by Theorem 5.5, we can equivalently consider bilipschitz images of k –regular trees for any k � 3.

6 Tilings of groups

In this section we prove that every countable group is poly-MT and every finitelygenerated group is poly-ccc. We refer the reader to the introduction for the relevantdefinitions and background.

We first consider poly-ccc groups. As the reader may notice, Corollary 5.7 plays acritical role in these proofs. It is unknown to the author how to prove these resultswithout applying Corollary 5.7.

Theorem 6.1 Every finitely generated group is poly-ccc.

Proof Let G be a finitely generated group. If G is finite then define T n D G and�n D f1Gg for all n 2 N . Then .�nIT n/n2N is a ccc sequence of monotilings.


206 Brandon Seward

Thus G is ccc and in particular poly-ccc. Now suppose that G is infinite. Let S be afinite generating set for G, and let d be the corresponding left-invariant word lengthmetric. For m> 1 let Fm be the nonabelian free group of rank m, and let F1 D Z bethe group of integers. By Corollary 5.7 there is a transitive translation-like action �of Fm on G for some m� 1.

In [6], Gao, Jackson and Seward prove that Fm is a ccc group. Thus there is a cccsequence of monotilings .�nIT n/n2N of Fm . Define a map �W Fm!G by setting�.t/D 1G � t . Since Fm acts on G transitively and translation-like, � is a bijection.Furthermore, if � is the standard left-invariant word length metric on Fm (associated tothe standard generating set of Fm ), then �W .Fm; �/! .G; d/ is Lipschitz (Lemma 5.1).Say the Lipschitz constant is c so that

d.�.t1/; �.t2//� c � �.t1; t2/

for all t1; t2 2 Fm . Now fix n 2N , and consider the family of sets

f�.ı/�1�.ıT n/ j ı 2�ng:

We have that for every t 2 T n ,

d.1G ; �.ı/�1�.ıt//D d.�.ı/; �.ıt//� c � �.ı; ıt/D c � �.1Fm

; t/:

Therefore the family of sets f�.ı/�1�.ıT n/ j ı 2�ng all lie within a common ball offinite radius about 1G 2G. Since balls of finite radius in .G; d/ are finite, the familyof sets is finite, say

f�.ı/�1�.ıT n/ j ı 2�ng D fT n

1 ;Tn2 ; : : : ;T

nk.n/g:

Notice that since � is injective, all of the T ni have the same number of elements.

Recall that 1Fm2 �n . By reordering the indices if necessary, we may assume that

T n1D �.1Fm

/�1�.1FmT n/D �.T n/. Now define for each 1� i � k.n/,

�ni D f�.ı/ j ı 2�

n and �.ı/�1�.ıT n/D T ni g:

Notice that 1G 2�n1

. Also notice that if ı 2�n and �.ı/ 2�ni , then

�.ı/T ni D �.ıT

n/:

The monotiling .�nIT n/ induces a partition of Fm , and this partition is mappedforward by � to a partition of G. The above expression shows us that the classes ofthis partition of G are precisely the left �n

i translates of T ni for 1� i � k.n/. Thus

Pn D .�n1; : : : ; �n

k.n/IT n

1; : : : ;T n

k.n// is a fair polytiling of G.



For n 2N the partition of Fm induced by .�nIT n/ is finer than the partition inducedby .�nC1;T nC1/, and thus the image of the first partition is finer than the image ofthe second. The images of these partitions are precisely the partitions of G inducedby Pn and PnC1 , respectively. So Pn induces a finer partition than PnC1 . Thusthe sequence .Pn/n2N is coherent. Also, as previously mentioned, 1G 2 �

n1

andT n

1D �.T n/� �.T nC1/D T nC1

1for all n2N . Since � is bijective and FmD

ST n ,

we have that G DS

T n1

. Thus .Pn/n2N is a coherent, centered and cofinal sequenceof fair polytilings. We conclude that G is poly-ccc.

The above theorem shows that a very large class of groups are poly-ccc. However,we don’t know if all countable groups are poly-ccc. We know even less about whichgroups are ccc.

Problem 6.2 Is every countable group poly-ccc? Is every countable group ccc?

Corollary 6.3 Every countable group is poly-MT.

Proof Let G be a countable group, and let F � G be finite. Then H D hFi is afinitely generated group and is thus poly-ccc by the previous theorem. Thus there is afair sequence of ccc polytilings

.Pn/n2N D .�n1; : : : ; �

nk.n/IT

n1 ; : : : ;T

nk.n//n2N

of H . By definition, T n1� T nC1

1and H D

ST n

1. Thus there is n 2N with F � T n

1.

Fix this value of n. Let D be a complete set of representatives for the left cosets of H

in G. Specifically, DH D G and dH \ d 0H D ¿ for d ¤ d 0 2D . It is easy to seethat .D�n

1; : : : ;D�n

k.n/IT n

1; : : : ;T n

k.n// is a fair polytiling of G. Since F � T n

1, we

conclude that G is poly-MT.

Problem 6.4 Is every countable group MT?

Below we address what values jT1j D jT2j D � � � D jTk j one can get from fair polytiles.T1; : : : ;Tk/. Recall that a group is locally finite if every finite subset generates afinite subgroup. If G is a locally finite group and .T1; : : : ;Tk/ is a fair polytile of G,then jT1j must divide the order of H D h

SkiD1 Tii. This is because any set aTi � aH

must either be contained in H or be disjoint from H . So .T1; : : : ;Tk/ is a fair polytilefor the finite group H , and thus jT1j divides jH j. Thus we see that in the case oflocally finite groups there are restrictions on what the cardinality of the sets Ti can be.The following theorem shows that this is the only obstruction.


208 Brandon Seward

Theorem 6.5 Let G be a countable nonlocally finite group. Then for every n � 1

there is a fair polytile .T1; : : : ;Tk/ of G with jT1j D n. Furthermore, if F � G isfinite then for all sufficiently large n there is a fair polytile .T1; : : : ;Tk/ of G withF � T1 and jT1j D n.

Proof Let F � G be finite. Since G is not locally finite, there must exist a finiteset S �G with hSi infinite. Set H D hF [Si. So H is a finitely generated infinitegroup. By Corollary 5.7 there is a transitive translation-like action � of Fm on H

for some m � 1. Define �W Fm ! H by �.t/ D 1H � t , and let Cay.Fm/ be thecanonical Cayley graph of Fm . In [6], it is proven that if ˆ is a finite connectedsubgraph of Cay.Fm/ then V.ˆ/ is a monotile of Fm . In particular, for every n� 1

there is a monotile T of Fm with jT j D n. It also follows that for every finite setF 0 � Fm and for all sufficiently large n there is a monotile T of Fm with F 0 � T andjT j D n. The statement of this theorem follows by using F 0 D ��1.F /, then applyingthe construction appearing in the proof of Theorem 6.1, and finally using left cosetrepresentatives for left cosets of H in G as in the proof of Corollary 6.3.

We have shown that all countable groups are poly-MT and all finitely generated groupsare poly-ccc, so it seems quite possible that all countable groups are poly-ccc. However,there are other natural tiling properties one can consider which lie between ccc andpoly-ccc and lie between MT and poly-MT. We do not study these notions here, butwe define them below so that others may investigate them.

Definition 6.6 A countable group G is super poly-MT if for every finite F �G thereis a fair polytile .T1; : : : ;Tk/ satisfying F �

TTi .

Definition 6.7 A countable group G is super poly-ccc if G has a fair ccc sequenceof polytilings

.�n1; : : : ; �

nk.n/IT

n1 ; : : : ;T

nk.n//

such that for each n we haveT

T ni �

TT nC1

i and G DSn

Ti

T ni .

The following diagram clarifies the relationship between the six tiling properties dis-cussed in this paper:

ccc��

H) MT��

super poly-ccc��

H) super poly-MT��

poly-ccc H) poly-MT



The ccc property is the strongest of all of the mentioned tiling properties, and thereare currently no groups which are known to not be ccc. However, showing that groupsare ccc can be quite difficult, and so studying some of the other related notions in thediagram may be more fruitful.

Problem 6.8 Which groups are super poly-ccc?

Problem 6.9 Which groups are super poly-MT?

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Department of Mathematics, University of Michigan530 Church Street, Ann Arbor, MI 48109, USA

[email protected]

www-personal.umich.edu/~bseward

Proposed: Martin Bridson Received: 6 April 2011Seconded: Peter Teichner, Walter Neumann Revised: 4 November 2011

Geometry & Topology Publications, an imprint of mathematical sciences publishers msp


http://dx.doi.org/10.1215/S0012-7094-99-09904-0


mailto:[email protected]

http://msp.org

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burnside's problem, spanning trees and tilings

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